General announcements• Today:

– Quiz 4 on rotational energy considerations, then introducing angular momentum

• Last quiz on angular momentum is Wednesday, after which we’ll wrap up loose ends on this unit and start talking about the unit test

• Friday or Monday (depending upon your period) we will review the test format and do a goalless problem in preparation for the second half of your block day

What’s left?• We’ve dealt so far with the rotational counterparts to kinematics, N2L, and

energy. Now: momentum!• A refresher:

– We know a net force accelerates a body. Similarly, a net torque angularly accelerates a body.

– Translational momentum depends on mass and velocity (p = mv)– A force applied for a certain duration of time can change momentum.

This is called impulse (𝐹∆𝑡 = 𝑚∆𝑣)

• Angular momentum is the rotational counterpart to momentum. A large angular momentum means it will take a relatively large torque applied over a given amount of time to bring a rotating body to rest.

1.)

Change your orientation, you change your motion . . . (courtesy of Yulia Lipnitskaya

The Island Series:

2.)

You have been kidnapped by a crazed physics nerd and left on an island with twenty-four hours to solve the following problem. Solve the problem and you get to leave. Don’t solve the problem and you don’t.

The problem: You are on a street walking a bicycle toward a curb. You want to lift the bike by its seat up and over the curb without having the front tire flopping around as you do it. Your task is to divine a maneuver that will insure the wheel stays straight.

Just before you get to the curb, accelerate the bike so the front wheel is rotating as fast as possible. The front wheel’s large angular momentum vector,directed along the wheel’s axle, will keep the wheel orientated as set. That is, as long as there are no external torques acting, the wheel will keep its orientation just as a gyroscope keeps its.

Solution to Island Problem

3.)

angular momentum

v

Angular momentumTranslational motion: Rotational motion:

Likewise, if all the torques acting on a group of particles (or just one) are zero or are internal to the system, then angular momentum is said to be conserved.

momentum angular momentum

For momentum, if all the forces acting on a group of particles are zero or are internal to the system, momentum is said to be conserved.

There is a lot here. To make life easier, we are going to go after these ideas in pieces.

6.)

Example 1: A point mass m moves with velocity v in a circular path of radius R. Determine its angular momentum using:

a.) Translational parameters:

b.) Rotational parameters:

Note: the direction of the angular momentum is perpendicular to the plane of the motion ( direction), as would be expected of a counterclockwise rotation.+k̂

Ipt mass = mR2

ω = vR

= R( ) mv( )sin90o

= mvR

!L = !r x !p

L = Iω

= mR2( ) vR

⎛⎝⎜

⎞⎠⎟

= mvR

!L

!ω

R mv•

Note the direction of the cross product!

Angular momentum example (MsDunham’s version)

Example: A point mass m circles a fixed point at a distance R units out. If its velocity is v, use both angular momentum relationships to determine the body’s angular momentum.

or

Shazam! It doesn’t matter which approach you use, you get the same value for the body’s angular momentum.

Changing angular momentumA few units ago, we started with Newton’s 2nd Law and did some rearranging to come up with the relationship

𝜏)*+ =∆𝐿∆𝑡 𝜏∆𝑡 = ∆𝐿 ∑𝐿. + ∑𝜏*0+∆𝑡 = ∑𝐿1

So, for rotational motion:

From that, we got the Conservation of Momentum relationship:

Important note about LWhat is different between the two conservation relationships is that for momentum, it is common to have an external forces acting over a time interval, whereas for angular momentum, it rarely happens that an external torque acts. In other words, with no external torques acting on a system, the conservation of angular momentum relationship usually ends up looking like:

One other thing: whereas rotational and translational kinetic energies can be put together in the same equation (both have units of Joules), angular momentum and translational momentum are NOT combinable! They have to be treated separately, like force and torque. Another way to remember:

– Translational momentum p has units 𝑁 3 𝑠 or 𝑘𝑔 3 𝑚/𝑠– Rotational momentum L has units 𝑁 3 𝑚 3 𝑠 or 𝑘𝑔 3 𝑚8/𝑠

L1∑ + τextΔt∑ = L2∑0

12.)

Example 4: An ice skater with arms out has an angular speed of and a moment of inertia . She pulls her arms in.

a.) What happens to her moment of inertia as she pulls her arms in? I1 I2

ω1

II

(it decreases)

b.) What is her new angular momentum?(no external torques, so it doesn’t change)

c.) What was her new angular speed?

There are no external torques acting on the woman, so conservation of angular momentum yields:

L1 + ΓextΔt = L2∑∑∑ I1ω1 + 0 = I2ω2

⇒ ω2 = I1ω1

I2

0

before after

. . . And as , her angular velocity increases.I2 < I1

But , so and mechanical energy is NOT conserved. This should not be a surprise. When the moment of inertia diminishes and angular velocity gets proportionally LARGER due to conservation of angular momentum, KE must go UP as it is governed by velocity (remember, ). 13.)

Example 4 (cont’) : An ice skater with arms out has an angular speed of and a moment of inertia . She pulls her arms in.

d.) Is mechanical energy conserved during this action? Justify and comment. I1 I2

ω1

II

Just by using your head, chemical energy in your muscles must be burned to force your arms inward, so you might expect that the mechanical energy in the system would not be conserved. Looking at the math, though:

Eo = 12

I1 ω1( )2 for the initial mechanical energy

before after

E2 = 12

I2 ω2( )2

= 12

I2I1

I2

ω1⎛⎝⎜

⎞⎠⎟

2

= 12

I2I1

2

I22 ω1

2⎛⎝⎜

⎞⎠⎟= 1

2I1ω1

2⎛⎝⎜

⎞⎠⎟

I1

I2

⎛⎝⎜

⎞⎠⎟= EO

I1

I2

⎛⎝⎜

⎞⎠⎟

I1I2>1 E2 > Eo

12Iω

2

Figure skater example (Ms. Dunham)http://ffden-2.phys.uaf.edu/w

ebproj/211_fall_2014/Ariel_Ellison/A

riel_Ellison/Angular.htm

l

There are no external torques acting on the woman, so although her angular momentum and speed will change, her angular momentum will not. Mathematically, this comes out to:

∑𝐿. + ∑ 𝜏*0+∆𝑡 = ∑ 𝐿1

𝐼:𝜔: + 0 = 𝐼8𝜔8⇒ 𝜔8 =

𝐼:𝐼8𝜔:

Note: her angular speed went up with moment of inertia going down so her angular momentum stays the same, BUT because angular speed governs energy (𝐾 = :

8𝐼𝜔8), the mechanical energy in the system goes UP. (This is due to chemical

energy burned in her body as she pulls her arms inward.)

Example: An ice skater with arms out have an angular speed of 𝛚1 and a moment of inertia I1. She pulls her arms in so her moment of inertia diminishes to I2. What happens to her angular speed?

14.)

•kid

kidwalk

walk

As they walk inward, each kid applies a force, hence torque, to the m.g.r., that changes the m.g.r.’s angular velocity. As a Newton’s Third Law action/action pair, the m.g.r. applies a force, hence torque, to the kids changing their angular velocity. As these are all internal impulses, conservation of angular momentum is applicable.

Example 5: Another standard problem is the merry-go-round problem. A merry-go-round, assumed to be a disk, has mass M and radius R. It also has two kids who push the m.g.r.’s outer edge by running along side of it to get it up to an angular speed of . The kids, each of which have a mass of , then jump on and start to walk toward the center of the m.g.r. When they get to within units from the center, they stop. What is their speed at that point?

ω1

mk

R3

Interestingly, there are TWO ways we can go here with the kids:

15.)

treating the two kids as masses executing rotational motion:

L1 + ΓextΔt = L2∑∑∑2 Ikid,1 ω1 + Imgr ω1( ) + 0 = 2 Ikid,2 ω2 + Imgr ω2( )

2 mkR2( )ω1 +

12

MR2⎛⎝⎜

⎞⎠⎟ ω1

⎛⎝⎜

⎞⎠⎟

+ 0 = 2 mkR3

⎛⎝⎜

⎞⎠⎟

2⎛

⎝⎜⎞

⎠⎟ω2 +

12

MR2⎛⎝⎜

⎞⎠⎟ ω2

⎛

⎝⎜

⎞

⎠⎟

⇒ ω2 = 2mk + M

2 2mk

9 + M2ω1 =

18 4mk + M( ) 4mk + 9M

ω1

the kid’s velocities:

v = R3

⎛⎝⎜

⎞⎠⎟ ω2 =

18 4mk + M( ) 4mk + 9M( )

R3

⎛⎝⎜

⎞⎠⎟ ω1

= 6 4mk + M( ) 4mk + 9M( )Rω1

6

16.)

treating the kids as point masses moving with velocity “v”:

L1 + ΓextΔt = L2∑∑∑2 !r1x

!p1 + Imgr ω1( ) + 0 = 2 !r2x!p2 + Imgr ω2( )

2 mkv1R( ) + 12

MR2⎛⎝⎜

⎞⎠⎟ ω1

⎛⎝⎜

⎞⎠⎟

+ 0 = 2 mkv2R3

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟+ 1

2MR2⎛

⎝⎜⎞⎠⎟

v2

R3( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

2 mk Rω1( )R( ) + 12

MR2⎛⎝⎜

⎞⎠⎟ ω1

⎛⎝⎜

⎞⎠⎟

+ 0 = 2 mkv2R3

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟+ 1

2MR2⎛

⎝⎜⎞⎠⎟

v2

R3( )

⎛

⎝⎜⎜

⎞

⎠⎟⎟

⇒ v2 = 2mk + M

2 2mk

3 + 3M2

Rω1= 6 4mk + M( ) 4mk + 9M

Rω1

Same solution either way. What’s important is the set-up, not all the nasty math.

MGR example – with a twist!A child of mass “m” runs with velocity “vi” tangent to a stationary merry-go-round and jumps onto it. The merry-go-round’s mass is M, radius R and moment of inertia ½ MR2 .

before jump after jump

mv

ω

MGR initially at rest

a.) Are there any torques acting on the system? If so, where do they happen? Are they internal or external?b.) is momentum conserved in the system?c.) is energy conserved in the system?d.) is angular momentum conserved in the system?e.) What’s the final angular velocity of the merry go round?

MGR with a twist (con’t.)!a.) Are there any torques acting on the system? If so, where do they happen? Are they internal or external?

There is a torque provided by the kid on the MGR when she jumps on. However, the MGR exerts an equal torque back on the kid, so these are internal torques.

No – there is an external forces providing an impulse during the collision due to the axle’s interaction with the ground

No – the kid is crashing into the mgr . . . it’s a perfectly inelastic collision

The torques are internal, so theory says yes, but there doesn’t seems to be any rotation and associated angular momentum before the collision, so this is confusing. How might we reconcile these two seemingly disparate observation?

b.) Is momentum conserved in the system?

c.) Is energy conserved in the system?

d.) Is angular momentum conserved in the system?

= mv(b)

!L = !r x !p = r2 (mv)sinθ2 = mv(r2 sinθ2 )

= mv b( )

!L = !r x !p = r1(mv)sinθ1 = mv(r1 sinθ1)

9.)

This is important: Objects moving in a straight line with constant speed HAVE angular momentum, and the angular momentum is CONSTANT!

This seemingly insane bit of amusement is actually grounded in intuitively sound reasoning. To see how, consider the simplest motion possible, a mass moving with constant velocity parallel to the x-axis. How does the angular momentum calculate out at several points in that case?

θ1r1 sinθ1 = b

r1θ1

θ2r2 sinθ2 = b

θ2r2

Same angular momentum! But how can a body moving STRAIGHT have angular momentum?

mv mv

A little bit later . . .

p⊥

Consider a body moving in circular motion. Its angular momentum will equal:

θ

L = !r x !p = p⊥r

p⊥

pradialθ

!r

p = mvθwhere is the body’s momentum. This momentum will be perpendicular to the position vector and tangent to the path. Nobody would argue that this body’s motion didn’t have angular momentum, as its motion is circular!

p⊥

Now consider a particle moving parallel to the x-axis with momentum mv, as shown in the sketch.

10.)

In other words, one of its components will be exactly like the momentum involved in the object that was executing a pure rotation, that had angular momentum. Conclusion, this body, moving in a straight line, will also have angular momentum(assuming its velocity vector’s line doesn’t go thru the reference point).

!r p⊥

It will have a momentum component that is radial and outward from the origin, and a tangential component ( ) that is perpendicular to the .

Back to the MGR…before jump after jump

mv

ω

MGR initially at rest

mvnew = m(Rω)

The kid’s initial momentum was mvi. After she jumps on, it’s mvnew = m(R𝛚). This means we can say:

e.) what is the final angular velocity?

∑𝐿. + ∑ 𝜏*0+∆𝑡 = ∑ 𝐿1

𝑅 𝑚𝑣. 𝑠𝑖𝑛90 + 0 = 𝑅 𝑚𝑅𝜔1 + 𝐼𝜔1

𝑅𝑚𝑣. = 𝑚𝑅8𝜔1 +12𝑀𝑅8𝜔1

𝜔1 =𝑚𝑣.

(𝑚𝑅 + 12𝑀𝑅)

Alternately, you could say after the collision, we have one object with total I = IMGR + Ikid as pt mass = ½ MR2 + mR2 so:

∑𝐿. + ∑ 𝜏*0+∆𝑡 = ∑𝐿1

𝑅 𝑚𝑣. 𝑠𝑖𝑛90 + 0 =12𝑀 +𝑚 𝑅8𝜔1

𝜔1 =𝑚𝑣.

(𝑚𝑅 + 12𝑀𝑅)

Angular momentum reminders• Angular momentum is the rotational version of momentum:

• If there are no external torques in a system (only internal torques), angular momentum is conserved! – In almost every situation, this will be true, so:

• Can you conserve angular momentum but not energy? Give an example to support your answer.Yes: figure skater (brings arms in, I decreases so angular velocity increases; rotational energy increases due to burning chemical energy in her arms)

Yes: merry go round (people walked from outside to inside; net I decreased so 𝛚increased; energy of system increased due to same reason as above)

à remember for this one, we had to consider the total I of the system before and after – the kids were point masses!

Another example: ball and beamSo with this in mind, consider a stationary meter stick of mass M (Icm = (1/12) ML2) sitting on a frictionless table. A puck of mass mp moving with velocity v0 hits the stick at its center of mass. What is conserved during the collision?

Energy: As is the case with all collisions, energy is not conserved here unless you are told the collision is elastic, which it isn’t in this case.

Momentum: There are no external forces, so momentum is conserved. The math:

Angular momentum: There are no external torques, so angular momentum, which is initially ZERO, is conserved. Since the ball hits the center of mass, the beam merely translates instead of rotating (no internal torque, either!)

Energy

Momentum:

Angular momentum:

Ball and beam v.2Now, the same collision but with the puck hitting a distance y units from the stick’s center of mass. What is conserved now?

Energy: Still no!

Momentum: Still yes, with:

Angular momentum: Still yes, except there is

L1 + ΓextΔt = L2∑∑∑⇒ mpvoy + 0 = −mpv1y + Istick,cmω

angular momentum about the stick’s center of mass due to the puck’s straight-line motion. That angular momentum equation becomes:

Energy:

Momentum:

Angular momentum:

One important (and strange) point:What’s unfortunate, at least from the perspective of problem solving, is that the velocity of the center of mass of the free stick is NOT related to the angular velocity about the stick’s center of mass by

That’s why you usually see this problem in the following alternate form: The stick is pinned at its center of mass, and when the puck hits, it stops dead. In that case:

v2 = rω = yω

Momentum is no longer conserved (an external force acts at the pin) but angular momentum is conserved with the final angular momentum being that of the stick only (the puck is not moving). With that, the stick’s final angular velocity can be calculated using the conservation of angular momentum equation, or:

General informationLast rotational motion quiz is tomorrow (Wednesday) on angular momentum

You will definitely see some version of the merry-go-round problem. Be able to:• apply concepts of conservation of angular momentum for a MGR with things moving

on its surface (e.g. kids moving from outside in or inside out, or jumping on); • Turn a translational momentum vector into a rotational momentum vector (r x p);• Understand what would cause a change in angular momentum (impulse due to

torque);• Understand whether momentum, angular momentum and/or energy are conserved;– There is also a good chance you will see some version of the meterstick/puck

problem (today’s) – be able to tell whether energy, momentum, and angular momentum are conserved for either situation (pinned or not).

• After the quiz, we’ll tie up some loose ends with angular momentum (fun examples - and The Wheel, pulsars, etc.) and start talking about the unit test + goalless problem for next week’s Block Day– Goalless problem guidelines are posted, but we’ll talk more Friday and I’ll have a

couple of practice ones for you

27.)

This is the rotational analogue to the problem shown to the right: A mass m sitting at the top of a curved incline of radius R slides down the incline,

Example 10: A mass m sits at the top of a curved, frictionless incline of radius R. It slides down the incline and executes a perfectly inelastic collision with the end of a pinned rod of mass 5mand length d. The two rotate up to some angle before coming to rest. If R = .4d, derive an expression for . You know:

θ

θ

m, d, R, g, and Icm,rod =5

12md2

i

m

d θ

5m

m

executes a perfectly inelastic collision with a 5m mass, and proceeds up a ramp. How high up the ramp does it go?

So how would you do this problem?(Energy up to the collision, momentum through the collision, energy after the collision!)

28.)

i

m

d θ

5m

How are the problems different as far as solving goes? That is, why can’t we just use conservation of momentum when the two masses collide in the pinned beam problem?

To use conservation of momentum, weneed a system in which there are no external impulses. The pin provides an external impulse (it keeps the rod from accelerating en-mass to the left through the collision), so conservation of momentum won’t work for this collision. There are no external torques acting about the pin, though, so conservation of angular momentum IS applicable.

m, d, R, g, and Icm,rod =5

12md2

29.)

i

m

To begin, the velocity of the mass m just before the collision can be determine using conservation of energy:

KE1∑ + U1∑ + Wext∑ = KE2∑ + U2∑ 0 + mgR + 0 = 1

2mv1

2 + 0

⇒ v1 = 2gR( )12

v1

v1

Because there are no external torques acting about the pin, conservation of angular momentum is the key to the collision. Taking a time interval through the collision, and summing the angular momenta about the pin, we can write:

L1,pin∑ + τextΔt∑ = L2,pin∑ L1,mass + 0 = L2,mass + L2,rod ⇒ !rx!p1 = Imassω2 + Ipinω2

We need those angular momentum quantities.

i

m

5m

m, d, R, g, and Icm,rod =5

12md2

30.)

To determine the angular momentum of the mass about the pin, we can go two ways. We can either treat the mass as a translating point mass and using , or we can use rotational parameters. I’ll show both (assuming the velocity of the mass at the bottom of the incline is ):

L1,m = !rx!p1

= d mv1( )

!rx!p

translating point mass:

L1,m = Imass/pin ω1

= md2( ) v1

d⎛⎝⎜

⎞⎠⎟

= md( )v1

rotational parameters:

Same either way.

v1

After the collision, the mass’s angular momentum in terms of angular velocity:

L2,m = !rx!p2

= d mv2( ) = d m dω2( )( ) = md2ω2

OR

i

ω1 =v1d

d

v1

Imass/pin = md2

m, d, R, g, and Icm,rod =5

12md2

31.)

We need the moment of inertia of the rod about the pin. We’ll use the parallel axis theorem for that:

Icm = Icm + Md2

= 512md2 + 5m( ) d

2( )2

= 53

md2

i

m

d θ

5m

Putting everything together through the collision yields:

L1,pin∑ + τextΔt∑ = L2,pin∑ !rx!p1,mass 0 = Imassω2 + Ipinω2

⇒ mdv1 = md2( )ω2 +53

md2⎛⎝⎜

⎞⎠⎟ ω2

⇒ ω2 =v1

d + 53d

= 3v1

8d

m, d, R, g, and Icm,rod =5

12md2

ω2 =v2

d, v2 = ω2d = 3v1

8d⎛⎝⎜

⎞⎠⎟ d

As

⇒ v2 =38

v1 =38

2gR

32.)

Knowing the after-collision velocities, we can use conservation of energy to determine how high the rod’s center of mass rises, and how high up the mass rises, before coming to a stop. Without doing the math to its conclusion, assuming the time interval is from just after the collision to thestop point and noting that the mass rises a distance (you should understand why by now) while the rod’s center of mass rises that equation looks like:

i

m

d θ

5m

m, d, R, g, and Icm,rod =5

12md2

KE1∑ + U1∑ + Wext∑ = KE2∑ + U2∑ KErod + KEmass( ) + 0 + 0 = 0 + Umass + Urod ( )12

Irod/pinω22 + 1

2mv2

2⎛⎝⎜

⎞⎠⎟ + 0 + 0 = 0 + mg d − dcosθ( ) + 5m( )g d

2− d

2cosθ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

d − dcosθ( )d2 −d2cosθ( ),

Δy = d − dcosθ